1
$\begingroup$

Young's modulus is typiccaly denoted with E. I have come accross a similar quantity called longitudal deformation modulus (in Russian literature, they call it модуль продольных деформаций) and it is denoted D. What is that quantity and is it the same as Young's modulus?

$\endgroup$
1
  • 1
    $\begingroup$ Show us some equations using your D. Then non-Russian-speaking engineers might be able to answer the question. $\endgroup$
    – alephzero
    Aug 6, 2018 at 0:42

1 Answer 1

1
$\begingroup$

Young's modulus is the elastic constant for a bar of a finite thickness which is free to contract or expand laterally as it is stretched lengthwise. In terms of the Lame' constants $\lambda$ and $\mu$ (see https://en.wikipedia.org/wiki/Lamé_parameters) we have $$ E= \lambda(1-2\sigma)+2\mu=\mu \frac{3\lambda+2\mu}{\lambda+\mu} $$ where $$ \sigma=\frac 12 \frac \lambda{\lambda+\mu} $$ is Poisson's ratio --- the ratio of lateral contraction to lengthways stretching. The elastic constant for longitudinal stretching of a sytem that is not allowed to shrink sideways as it is stretched - for example the longitudinal "p" waves in an earthquake --- is different. It is given by $\lambda+2\mu$, so the velocity of longitudinal $p$-waves is $$ V_p = \sqrt{\frac{\lambda+2\mu}{\rho}} $$ where $\rho$ is the density.

$\endgroup$
2
  • $\begingroup$ Then if the elastic constant for longitudinal stretching is the same as longitudal deformation modulus, it should not be easily possible to measure it directly but typically only from the p-wave velocity. But the paper I read said they measured the value by electrotensometry during uniaxial compression (where the sample is allowed to change in tranversal directions). $\endgroup$
    – atapaka
    Aug 5, 2018 at 22:34
  • $\begingroup$ If this is what the Russian text means, then the longitudinal modulus is related to $E$ and $\nu$ both of which are easy to measure in a uniaxial compression (or tension) test. The relationship would be something like $D = \dfrac{E}{1-\nu^2}$ - if this really is the meaning of $D$, of course. $\endgroup$
    – alephzero
    Aug 6, 2018 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.